Posted by **SUPER HARD** on Sunday, November 25, 2012 at 8:21am.

In triangle ABC, m<A = 90, m<B=75, and BC = \sqrt{3}. What is the area of ABC? Express your answer as a common fraction.

HELP THIS IS SO HARD HHELP MEE!!! EXplain too please!

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**Reiny**, Sunday, November 25, 2012 at 12:38pm
You have a right-angled triangle so we can use the formula

(1/2) base x height

Since you want the answer as a common fraction, I will assume that you need "exact" values.

you state that BC = \sqrt{3}

I will assume that is meant to say √3

so cos75 =base/√3

base = √3cos75

and sin75 = height/√3

height = √3sin75

area = (1/2)(√3sin75)(√3cos75)

= (3/2)sin75cos75)

(recall sin 2A = 2sinAcosA)

= (3/2)(2sin75cos75)/2

= (3/4) sin 150

sin 150 = sin30 ---- one of the basic identities

area = (1/3) sin150

= (3/4)sin30

= (3/4)(1/2) = 3/8

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**SUPER HARD**, Sunday, November 25, 2012 at 3:46pm
thanks alot

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**BOB**, Wednesday, August 3, 2016 at 2:18pm
THANK YOU !!!!!!!

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**Ok OK**, Friday, August 26, 2016 at 7:02pm
You don't even need trig for this. Reflect B, draw and altitude from B to B'C and solve

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